How would you integrate ye^(6y-y^2) wrt y?

Answer 1

First, look at 6y-y² = -(y-3)²+9 by completing the square

e^{-(y-3)^2+9} = e⁹e^{-(y-3)^2}

Let u = y-3, du = dy, y=u+3

f(u) = (u+3)e⁹e^{-u^2}

Split into to integrals with respect to u

I=e⁹[∫ue^{-u^2}du + 3∫e^{-u^2}du]

It is known the ∫e^{-u^2}du is √πerf(u)/2, where erf(u) is the error function.

∫ue^{-u^2}du = -e^{-u^2}/2 using a substitution

Putting the two back together and setting u=y-3

I = e⁹[√πerf(y-3)/2-e^{-(y-3)^2}/2] + C